Answer:
[tex]UCL = 0.078[/tex]
[tex]LCL = 0[/tex]
Step-by-step explanation:
Poorly formatted question (see attachment)
From the attachment, we have:
[tex]n = 135[/tex] --- the sample size
[tex]Samples = 20[/tex] --- The number of samples
[tex]\sum np = 87[/tex] --- Number of leaking tubes
First, we calculate the number of observations
[tex]\sum n = n * Samples[/tex]
[tex]\sum n= 135 * 20[/tex]
[tex]\sum n= 2700[/tex]
Using 3 sigma control limit, we have:
[tex]z = 3[/tex]
Calculate [tex]\bar p[/tex]
[tex]\bar p = \frac{\sum np}{\sum n}[/tex]
So, we have:
[tex]\bar p = \frac{87}{2700}[/tex]
[tex]\bar p = 0.0322[/tex]
Next, calculate the standard deviation
[tex]s_p = \sqrt{\frac{\bar p* (1-\bar p)}{n}}[/tex]
So, we have:
[tex]s_p = \sqrt{\frac{0.0322* (1-0.0322)}{135}}[/tex]
[tex]s_p = \sqrt{\frac{0.03116}{135}}[/tex]
[tex]s_p = \sqrt{0.0002308}[/tex]
[tex]s_p = 0.0152[/tex]
The control limits is then calculated as:
[tex]UCL = \bar p + z * s_p[/tex] --- upper control limit
[tex]LCL = \bar p - z * s_p[/tex] --- lower control limits
So, we have:
[tex]UCL = \bar p + z * s_p[/tex]
[tex]UCL = 0.0322 +3 * 0.0152[/tex]
[tex]UCL = 0.078[/tex]
[tex]LCL = \bar p - z * s_p[/tex]
[tex]LCL = 0.0322 - 3 * 0.0152[/tex]
[tex]LCL = -0.013[/tex]
Since the calculated LCL is less than 0, we simply set it to 0
[tex]LCL = 0[/tex]
So, the p chart control limits are:
[tex]UCL = 0.078[/tex]
[tex]LCL = 0[/tex]
